I have the following problem: 70% of women respond positively to a test, while only 40% of men do so. If 10 participants are selected (5 women and 5 men), what is the probability that only 1 man responds negatively?
Now what I have done is to have two distributions, one for women and one for men. Then I calculated P(Y=5) for women with p=0.7 and P(Y=4) for men with p=0.4, and since the events are mutually exclusive I´ve added the two, having a result of ~24%. Can this be correct...?
On
You have the right idea, but did some wrong calculations.
The probability of 4 men responding positively and one man responding negatively is $_5C_4(0.4)^4(0.6)^1$.
The probability of 5 women responding positively is $_5C_5(0.7)^5(0.3)^0$.
The product of these two calculations is what you are looking for.
On
Almost!
You've correctly figured out how to calculate the probability of picking five women and four men answering positively: $$P_W(X=5)= {5 \choose 5} (0.7)^5 (0.3)^0 \approx .168$$ $$P_M(X=4)= {5 \choose 4} (0.4)^4 (0.6)^1 \approx .077$$
Your mistake was in adding the two. You require that both of these events occur. Since they are independent, you may simply multiply the probabilities of the events occurring individually to obtain this: $$.168 \times .077 \approx .013$$
To more clearly see why you cannot add the two, recalculate your answer in the case where women answer positively $100\%$ of the time. You'll get a likelihood of more than $100\%$!
The events of $5$ women out of $5$ respond positively and $4$ men out of $5$ respond positively are not mutually exclusive. In fact it is perhaps reasonable to assume that they are independent, and so you can multiply the probabilities.